If the velocity function is given by V(x) = 12x - 4x^2, find an equation of the tangent line to the graph of V(x) at (1, 8).

Understand the Problem

The question asks to find the equation of the tangent line to the graph of the velocity function V(x) = 12x - 4x^2 at the point (1, 8). This involves finding the derivative of V(x), evaluating it at x=1 to get the slope of the tangent line, and then using the point-slope form of a line to find the equation of the tangent line.

Answer

$y = 4x + 4$

Steps to Solve

Find the derivative of $V(x)$

To find the slope of the tangent line, we first need to find the derivative of the velocity function $V(x) = 12x - 4x^2$. We'll use the power rule.

$$ V'(x) = \frac{d}{dx}(12x - 4x^2) = 12 - 8x $$

Evaluate the derivative at $x=1$

Now, we evaluate the derivative at $x=1$ to find the slope of the tangent line at the point $(1, 8)$.

$$ V'(1) = 12 - 8(1) = 12 - 8 = 4 $$

So, the slope of the tangent line is $4$.

Use the point-slope form of a line

We have the point $(1, 8)$ and the slope $m = 4$. We can use the point-slope form of a line to find the equation of the tangent line:

$$ y - y_1 = m(x - x_1) $$

Plugging in the values:

$$ y - 8 = 4(x - 1) $$

Simplify the equation to slope-intercept form

Now, we simplify the equation to the slope-intercept form ($y = mx + b$).

$$ y - 8 = 4x - 4 $$ $$ y = 4x - 4 + 8 $$ $$ y = 4x + 4 $$

$y = 4x + 4$

More Information

The tangent line to the graph of $V(x) = 12x - 4x^2$ at the point (1, 8) is $y = 4x + 4$. This line has the same slope as the curve at the given point and passes through that point.

Tips

A common mistake is to forget to add the $y_1$ value after distributing $m$ ($y - y_1 = m(x - x_1)$). Another mistake is incorrectly calculating the derivative.

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